A) R
B) \[\left\{ n\pi \mp \frac{\pi }{3},n\in I \right\}\]
C) \[\left\{ n\pi \pm \frac{\pi }{6},n\in I \right\}\]
D) None of these
Correct Answer: C
Solution :
We have,\[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{(2\sin x)}^{2n}}}{{{3}^{n}}-{{(2\cos x)}^{2n}}}\] Clearly, \[f(x)\]is discontinuous when \[{{3}^{n}}-{{(2\cos x)}^{2n}}=0\] \[\Rightarrow \]\[{{(\sqrt{3})}^{2n}}-{{(2\cos x)}^{2n}}=0\] \[\Rightarrow \]\[{{(\sqrt{3})}^{2n}}=2{{(\cos x)}^{2n}}\] \[\Rightarrow \]\[\sqrt{3}=2\cos x\] \[\Rightarrow \]\[\cos x=\frac{\sqrt{3}}{2}\] \[\Rightarrow \] \[x=n\pi \pm \frac{\pi }{6},n\in I\]
You need to login to perform this action.
You will be redirected in
3 sec
